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In how many ways can a pack of $36$ cards be split in two equal parts so that each portion contains two aces (assume $4$ aces )

My Try :

We have $36$ cards and $4$ aces and we have split into $2$ equal part

so i used $32C2$ , which is wrong answer .

SSK
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2 Answers2

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There are two players, Alicia and Beti. We first count the number of ways to split the deck between Alicia and Beti. (This is a different question from yours.)

There are $\binom{4}{2}$ ways to choose the Aces to give to Alicia, and for each such choice there are $\binom{32}{16}$ ways to produce the rest of Alicia's hand, for a total of $\binom{4}{2}\binom{32}{16}$.

This counts the number of splittings (receiver unspecified) twice, once as Alicia's hand and once as Beti's. So for the answer to the actual problem, divide by $2$.

Another way: We want to split the cards into two teams. The Ace of Spades will choose her team.

She can choose the Ace that will keep her company in $3$ ways. Then she can choose the non-Aces on her team in $\binom{32}{16}$ ways.

André Nicolas
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  • How you found the $\binom{32}{16}$ – SSK Jul 01 '13 at 17:47
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    There are $32$ non-Aces, each hand will contain $16$. By $\binom{32}{16}$ I mean what in high school is sometimes called ${}{32}C{16}$, or something like that, the Combination symbol. Is that your question? – André Nicolas Jul 01 '13 at 17:50
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Hint:

Think of it this way:

First put the 2 aces in each pile (how many ways do we have to do that?) Then, how many cards do you have left to divide between the piles?

Nathaniel Bubis
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